GE1151 – ENGINEERING

MECHANICS (Common to all Branches)

CONCEPTS & DEFINITIONS

COMPILED BY

DEPT. OF MECH.

N I COLLEGE OF ENGINEERING

KUMARACOIL

GE1151 – ENGINEERING MECHANICS

(Common to all Branches)

CONCEPTS AND DEFINITIONS

MODULE I - BASICS AND STATICS OF PARTICLES

1. UNITS

QUANTITY mass time current temp intensity length

UNIT kilogram second ampere Kelvin candela Meter

2. PARELLOGRAM LAW

R=Sqrt (P2+Q2+2PQCOSa)

Where R�Resultant P�1stforce Q�2ndforce a�Angle between two forces

3. SCALAR QUANTITY

“It’s one having only magnitude”

4. VECTOR QUANTITY “It’s one having both magnitude and direction”

5.POSITION VECTOR

R=Sqrt(x2+y2+z2)

6. DOT PRODUCT

. I J k

I 1 0 0

j 0 1 0

k 0 0 1

7. CROSS PRODUCT

X I J k

I 0 K -j

j -k 0 I

k J -I 0

8, LAMIS THEOREM If three coplanar forces are acting at a point in equilibrium Then each force is directly proportional to sine of opposite two forces. P Q R = = SIN A SIN B SIN C 9. LAW OF MECHANICS 1. A particle remains in its position if resultant force is zero 2. If the resultant force is not zero then acceleration is proportional to resultant force F= ma. 3. Action and reaction forces are in same line of action, equal in equal in magnitude but opposite in direction. 10. NEWTONS LAW OF GRAVITATION The law states that two particles of mass `m` and mutually attracted By two opposite forces F,-F then magnitude of F is given by F = G Mm Newton R2 11. PRINCIPLEOF TRASMISSIBILITY

Principle of transmissibility states that “ the motion of a rigid body remains unchanged if a force acting on a point is replaced by another force having same magnitude and direction in the same line of action. 12. RESULTANT FOR MORE THAN 2 FORCES For a system when resolving Fh = F cosA Fv= F sin B The resultant force is

R=Sqrt ((έh) 2+ (έv) 2) 13. EQULIBRANT A force E with same magnitude and same line of action but opposite Direction is called equilibrant. 14. EQULIBRIUM The equilibrant used to arrest the movement of the particle, then the The body is said to be in equilibrium. Here Resultant = 0. R F1

F2 F4 F3 15. CONDITIONS FOR EQUILIBRIUM

For collinear forces £H = 0 , £V = 0 For concurrent forces £H = 0 , £V = 0 16. FREE BODY A body which has been isolated from surroundings is called Free body. 17. FREE BODY DIAGRAM The sketch showing all the forces and moments acting on the Free body is called free body diagram 18. HORIZONTAL & VERTICAL COMPONENTS IN EACH QUATRANT

1) TOWARDS ORGIN

F h Fv

1st -VE -VE 2nd +VE -VE 3rd +VE +VE 4th -VE +VE

2) OUTWARDS FROM ORGIN

F h Fv 1st -VE -VE 2nd +VE -VE 3rd +VE +VE 4th -VE +VE 19. FORCES I N THREE DIMENSION Consider a system of force in three dimensions So that _

F = F cosØx + F sinØy +FsinØy

Øx=cos-1 ( Fx / F )

Øy=cos-1 ( Fy / F )

Øy=cos-1 ( Fy / F )

20. CONDITIONS

COS2ØX + COS2ØY + COS2ØZ = 1

L2 + m2 + n2 = 1

21. RESULTANT

_ _ _ _ R = Rx I + Ry j + Rz k R = Sqrt ( Rx2 + Ry2 + Rz2 ) Where

Øx=cos-1 ( Rx / F )

Øy=cos-1 ( Ry / F )

Øy=cos-1 ( Ry / F )

22. POSITION VECTOR If Coordinates of A is ( X1 , Y2 , Z3 ) Coordinates of B is ( X2 , Y2 , Z3 ) Then __ Rab = x2 + y2 + z2

Where X = x2 - x1 Y= y2 - y1 Z= Z2 - z1 23. FORCE VECTOR

--- Fab = Fab. (UNIT VECTOR) where _ _ _ Unit vector = Fx I + Fy j + Fz k Sqrt ( Fx2 + Fy2 + Fz2) 24. EQUATION OF FORCE IN THREE DIMENSION

έFx =0

έFy =0

έFz =0

MODULE II - STATICS OF RIGID BODIES

1. Rigid bodies

When a body is not subjected to collinear or

concurrent force system, then the body is to be

idealized as a rigid body.

2. Moment of force

Moment of force is defined as the product of the force and the

perpendicular distance of the line of action of force from the

point. It’s unit is N-m.

M=F*r N-m

3. Clockwise moment → positive sign

Anticlockwise → negative sign

4. Varignon’s theorem of moment :

The algebraic sum of moments of any number of forces about

any point in their plane is equal to moment of their resultant

about the same point.

(ie.) F1d1 + F2d2+ F3d3 + - - - - - - - - = R × d

5. Moment passing through the reference point is zero.

6. Resultant of non-parallel , non-concurrent and coplanar forces is

given by,

R = √ (∑H)2 + (∑V)

2

where ∑H = Fh1 + Fh2 + Fh3 + - - - - - - - -

∑V = FV1 + FV2 + FV3 + - - - - - - -

θ = tan-1

(∑H / ∑V)

7. In the parallel force system, resultant should be in between the two forces and parallel to the forces.

8. Force – couple system

Couple:

Two forces F and –F having the same magnitude,

parallel line of action but in opposite direction are said to form

a couple.

9. For magnitude of the resultant force, magnitude of couple is not

considered.

10. In a couple, sum of the moments of two forces about any point is

same in magnitude and direction.

11. In general, moment of couple = F × arm length

12. Types of couple

a. Clockwise couple

b. Anticlockwise couple

13. Equilibrium of rigid body in two dimension

For the equilibrium state of rigid body, the resultant

force and the resultant force and the resultant moment with

respect to any point is zero.

∑H = 0 ; ∑V = 0 ; ∑M = 0

14. For parallel forces acting on a rigid body

∑V = 0 ; ∑M = 0

15. For inclined forces acting on a rigid body

∑V = 0 ; ∑H = 0 ; ∑M = 0

16. Difference between moment and couple

The couple is a pure turning effect, which may be moved

anywhere in it’s own plane whereas, moment of force must

include a description of the reference axis about which the

moment is taken.

17. Action

The self weight of a body acting vertically downwards is

known as action.

18. Reaction

The resulting force against the action acting vertically upwards

is known as reaction. It is developed at the support.

19. Support reaction

The resistance force on the beam against the applied load at

support is called support reaction.

20. Types of support

i. Roller support

ii. Hinged support

iii. Fixed support

21. Roller support has only one reaction

22. .Hinged support is also called pin-joint support.

23. Roller and hinged supports can resist only displacement but

rotation of beam is not resisted by both the supports.

24. Roller support has the known line of action i.e. always normal to

the plane of rollers.

25. Hinged support has an unknown line of action i.e. at any angle θ, with the horizontal

26. Vertical reactions are assumed upwards and horizontal reactions

outwards. While solving, if negative value is obtained, then the

direction is reversed.

27. Reaction RA= √ HA2 + VA

2

θ = tan-1

(VA/HA)

28. Fixed support

It has three reactions at a point.

29. Types of load on a beam

1. Point load

2. Uniformly distributed load (UDL)

3. Uniformly varying load (UVL)

30. Uniformly distributed load (UDL)

This load is applied all over the area uniformly. It acts exactly

at the mid-point of the body.

31. Uniformly varying load (UVL)

The total load acts at the centroid of the triangle.

32. If the beam is supported at the two ends, it is known as simply

supported beam (SS beam).

33. Supported reactions of trusses

Trusses are made up of several bars and joints. It is subjected

to point loads at the joints.

34. Neglect the self weight of the truss, if it is given in the problem.

35. If truss is given in a problem, modify the truss configuration to

beam configuration and then solve the problem.

36. To find the moment of force about an axis, the moment of the force

about any point, lying on the particular axis should be known.

37. Moment of force about a point is a vector quantity, where as the

moment of force about the axis passing through that point is a scalar

quantity.

38. If all the forces are in the XZ plane, parallel to Y-axis, then scalar

approach can be used.

MODULE-III- PROPERTIES OF SURFACES AND SOLIDS

1. CENTROID

CENTROID is the point where the entire weight of the body will act

in Two Dimensional. In this case mass of the body is neglected.

2. CENTER OF GRAVITY

CENTER OF GRAVITY is the point where entire weight of the

body will act in Three Dimensional. Here weight of the body is considered.

3. MOMENT OF INERTIA

i. MOMENT OF INERTIA about any axis in the plane may be

defined as the sum of the product of each elemental area and the

square of the perpendicular distance between the element and the

axis.

ii. M.I. is also called as SECOND MOMENT OF AN AREA.

iii. M.I. of an area about x-axis is represented as Ixx and about y-axis is

represented by Iyy.

iv. Unit of moment of inertia is mm4.

4. PERPENDICULAR AXIS THEOREM

The perpendicular axis theorem states that the moment of inertia of a

plane area about an axis perpendicular to the plane and passing through the

intersection of the other two axes xx and yy contained by the plane is equal to

the moment if inertia about xx and yy.

Izz = Ixx+ Iyy

5. POLAR MOMENT OF INERTIA:(IP)

The polar moment of inertia is the moment of inertia about the pole

(Ip) from the distance R.

IP=Ixx+Iyy

6. RADII OF GYRATION

If the plane area can be represented as plane figure parallel to xx axis

and parallel to yy axis .Kx is the distance from yy and Ky is the distance from

xx then

MI along x axis=Ixx = ∫y2 dA

Ixx = kx2A

Radius of Gyration Kx = √(Ixx/A)

MI along y axis=Iyy = ∫ x2 dA

= ky2A

Radius of Gyration Ky = √(Iyy/A)

7. PARALLEL AXIS THEOREM

The theorem states that the moment of inertia of the plane area about

any axis parallel the centroidal axis (Iaa) is equal to moment of inertia of the

area about the centroidal axis(ICG) added to product of area(A) and square of

the perpendicular distance between the two axes(h).

Iaa =ICG + Ah2

ICG is also called as self inertia.

8. PRODUCT OF MOMENT OF INERTIA

The product of inertia for the elementary area is defined as dIxy and is

equal to (xy) (dA). Thus, for the entire area A, the product of inertia is,

Ixy = ∫ xy dA

Ixy = A1x1’y1

’ +A2x2

’y2

’ +………+Anxn

’yn

‘

_

_

Where

X1

’= X-X1, X2

’= X-X2 ……

_ _

Y1’= Y-Y1, Y2’=Y-Y2…....

9. MASS MOMENT OF INERTIA

The property which measures the resistance of a body to angular

acceleration is the mass moment of inertia of the body.

10. STANDARD FORMULA FOR CENTER OF GRAVITY

CENTER OF GRAVITY

SHAPES

AREA

_

X

_

Y

Rectangle b×h b/2 h/2

Triangle 1/2*×b×h 2/3×b 1/3×h

Circle πd2/4 d/2 d/2

Semicircle πd2/8 d/2 2d/3π

Quarter circle πd2/16 4r/3 4r/3π

b= base length h=height of the section d=diameter r = radius

11. Center of gravity for compound section:

The center of gravity of compound sections formed by various

common shapes of areas and lines may be found by splitting the compound

sections into rectangles, triangles or other shapes.

_

X= (A1 X1 +A2 X2 +A3 X3…)/ (A1 +A2 +A3……)

_

Y= (A1 Y1 +A2 Y2 +A3 Y3…)/ (A1 +A2 +A3……)

_ _

X1, X2 X3……and Y1 Y2 Y3….. are X and Y values of

individual section. A1, A2, A3…..are the areas of individual sections.

12. MI of composite area:

IXX = Σ [(Iself) xx + Area × (distance) 2]

Iyy = Σ [(Iself) yy + Area × (distance) 2]

Iself= Self MI along xx/yy axis

_

For IXX , distance p=(y-y)

_

IYY , distance p=(x-x)

13. SELF MOMENT OF INETIA :

SHAPES SELF MOMENT OF INERTIA

Rectangle (Iself)XX =bd3/12

(Iself)YY=db3/12

Triangle (Iself)XX =bh3/36

(Iself)YY=hb3/36

Circle (Iself)XX =πd4/64

(Iself)YY=πd4/64

Semicircle (Iself)XX =0.11×r2

(Iself)YY=0.39×r4

Quarter circle (Iself)XX =0.055×r4

(Iself)YY=0.055×r4

14. PRINCIPAL MOMENT OF INERTIA:

tan2θ = -2IXX/ (IXX-IYY)

Imax/min= (IXX+IYY)/2± √(((Ixx-IYY) 2/2) +IXY)

15. MASS MOMENT OF INERTIA:

IAA, mass=∫r2ρ(tdA)

=ρt× IAA, area where IAA, area =∫r2dA

This is the relation between mass moment of inertia and area moment

and area moment of inertia.

16. Unit of mass moment of inertia is kg mm2.

17. DERIVATION OF MMI OF RECTANGULAR PLATE:

Consider a rectangular plate of width b depth d and thickness t

We have IAA, mass=ρt× IAA, area

=ρt ×(db3/12)

=ρtdb (b2/12)

= (mb2/12) [mass m=ρtdb]

IBB, mass=ρt× IBB, area

=ρt ×(bd3/12)

=ρtdb (d2/12)

= (md2/12) [mass m=ρtdb]

ICC, mass = IAA, mass+ IBB, mass

ICC, mass =m/12(b2+d

2)

18. DERIVATION OF MMI OF CIRCULAR PLATE:

Consider a thin circular plate of radius r

IAA, mass=ρt× IAA, area

=ρt (πr4/4)

=ρtπr2 [r

2/4]

= (mr2/4)

= IBB, mass

ICC, mass = IAA, mass+ IBB, mass

= (mr2/2)

19. The radius of gyration k may be defined as the distance at which the entire

mass of the body should be concentrated without altering the mass moment of inertia.

MODULE IV - DYNAMICS OF PARTICLES

1. DISPLACEMENT (S)

It is the distance travelled by the particle. It is a scalar quantity. Unit:

meter

2. VELOCIYT (V)

It is the rate of change of displacement. It is a vector quantity. Unit:

ms-1

V=ds/dt

3. ACCLERATION (a)

It is the rate of change of velocity. It is a vector quantity.Unit: ms-2

a = dv = d²s

dt dt

Acceleration = Change of velocity= V-U

Time taken t

4. RETARDATION / DECELERATION

If initial velocity > final velocity then the acceleration is

negative

5. SPEED

Speed = distance travelled

Time taken

6. AVERAGE SPEED

Average speed = total distance travelled

Total time taken

7. AVERAGE VELOCITY

Average velocity = (initial + final ) velocity

2

= u+v

2

8. DISTANCE TRAVELLD (S)

s = velocity x time

s = vt

9. EQUATIONS OF MOTION IN A STRAIGHT LINE

v=u+at

s=ut+½at²

v²=u²+2as

u = 0, body starts from rest

v = 0 , body comes to rest

10.1 Km/Hr = 0.277 m/s = 1000 m/s

3600

12. DISTANCE TRAVELLED IN N SECONDS

Distance travelled in n

th second = sn - sn

sn > sn-1

s nth

= u +a (2n – 1)

2

13. EQUATION OF MOTION OF PARTICLE UNDER GRAVITY

when particle comes down:

h = ut + gt²

v = u + gt

v² = u² + 2gh

when particle moves upwards :

h = ut - gt²

v = u - gt

v² = u² - 2gh

14. MAXIMUM HEIGHT ATTAINED BY PROJECTED

PARTICLE

(V=0)

h max = u²

2g

15. TIME TAKEN TO REACH MAX. HEIGHT

(V = 0)

t = u

g

16. TOTAL TIME TAKEN

Total time taken = 2 x time taken to reach

max. Height

= 2u

g

17. STRIKING VEIOCITY OF THE PARTICLE

(U = 0)

v = (2gh) ½

Velocity with which a particle is thrown (v = 0)

u = (2gh) ½

18. DISPLACEMENT IN TERMS OF TIME

s = f (t)

v = ds = d f (t)

dt dt

a = dv = d² f (t)

dt dt²

19. EQUATION OF MAXIMUM VEIOCITY

Equation of maximum velocity dv = 0

dt

20. RECTILINEAR MOTION WITH VARIABLE ACCELERATION

a = f (t)

⇒ v = § f (t) dt

s = §§ f (t) dt

= § v dt

a = v dv

ds

21. RELATIVE VELOCITY: � VB/A =VB – VA Here VB>VA

� VA/B = VA – VB Here VA>VB

� VB/A = (VB² + VA²)½

� θ = tan-1

VB

VA

22. RELATIVE DISPLACEMENT:

S B/A = SB - SA

S B/A = (SA² + SB²)½

θ = tan-1 SB

SA

23. RELATIVE ACCELERATION:

a B/A = (aA² + aB²)½ θ = tan

-1 aB

aA

24. KINEMTICS OF PARTICLES – CURVILINEAR MOTION:

VEIOCITY:

Horizontal component, Vx = dx

dt

Vertical component, Vy = dy

dt

Resultant velocity, v = (vx²+vy²)½

Angle of inclination with x – axis α = tan ¯1 vy

vx

25. ACCELERATION:

a = (ax² + ay²)½

θ = tan¯1

ay

ax

Horizandal Component:

ax = dvx

dt

Vertical Component: ay = dvy

dt

26. NORMAL ACCELERATION:

an = v²

r

v = velocity of the particle

r = radius of the curvature of curvilinear

motion

27. RESULTANT ACCELERATION:

a = (at² +an²)½

at = tangential acceleration

an = normal acceleration

POLAR SYSTEM (r,θθθθ)

28. VELOCITY COMPONENT:

radial component Vr = dr

dt

Transverse component Vθ = r dθ

dt

V = (Vr² + Vθ²)½

θ = tan¯ 1 V r Vθ

29. ACCELERATION COMPONENT: Radial component ar = d² r -r dθ ²

dt² dt

Transverse component aθ = rd²θ + 2dv dθ

dt dt dt

30. PROJECTILE:

A Particle projected in the space at an angle to

horizandal. Plane is called projected. And motion is called projectile motion.

31. TIME OF FLIGHT:

t = u sinα

g

u sinα = initial velocity in the upward (vertical

component)

α = angel of projection

total time T = 2 u sinα

g

32. MAXIMUM HEIGHT ATTAINED:

h max = u²sin²α

2g

33. RANGE, (R)

R = u²sin²α

2g

Maximum horizontal range, R = u²

g

when α = 45 0

34. VELOCITY AND DIRECTION OF THE PROJECTIE

Vx = u cos α

Vy= (u sinα)²-2gh ½

V= (Vx²+Vy) ½

35. MOTION OF PARTICLE FROM KNOWN HEIGHT

Range=Horizandal Velocity ×time taken

= u x t

h = gt²

2

36 . WORK DONE Work Done=Force X displacement

Unit=Nm=Joules

37. POWER Rate of doing work

Power = Work done

Time

=Force x displacement

Time

= force x velocity

=unit: Nms¯1 = watt

NEWTONS LAWS OF MOTION

38. MOMENTUM Momentum= Mass x Velocity

M=mv

Unit: Kgm¯1

1 Newton=kgm¯1

39. WEIGHT

Weight=mass x acceleration due to gravity

W = mg

Unit: Newton

40. STATIC EQUATION OF EQULIBRIUM

∑H=0; ∑V=0; ∑M=0

41. D’ALEMBERTS EQUATIONS OF EQILIBRIUM

P-ma= 0

a = P

m

a= P1+P2+P3 ……...

m

D’alelemberts principle stats that system of force acting on a body in

motion is in Equlibrium with the inertia a force or imaginary force (ma) of the

body

42. MECHANICAL ENERGY:

P.E = mg x h

= (Force) x (Displacement)

43. KINETIC ENERGY:

KE = ½mv² [if u = 0]

= ½m (v² - u²) [u ≠ 0]

= w (v² - u²)

2g

= ½mv² - ½mu²

Work done by body in motion = final kinetic energy _ initial kinetic

energy

∑fx = sum of forces that induce the motion of a body.

44. WORK DONE BY SPRING BODY:

Work done = -½ kx²

Unit: N/mm = KN/m

K ⇒ spring constant / spring modulus (depends upon the material

used)

K = f , x = maximum deformation

x

f = kx

45. DEFORMATION TO NORMAL POSITION:

Work done = ½kx²

Unit: N/mm

46. WORK DONE BY A SYSTEM:

= work done by the block + work done by

the spring

47. SPRING CONSTANT:

= load.

Deflection

Unit: N/m

IMPULSE – MOMENTUM

48. IMPULSE OF A FORCE When a large force acts for a short period of that force is

called an impulsive force. It is a vector quantity

Impulse (I) = force x time

Unit = Ns

49. MOMENTUM Momentum = mass x velocity

M = mv

Unit = Kgms¯1

50. IMPULSE-MOMENTUM EQUATION Impulse = Final momentum-Initial momentum

51. LAW OF CONSERVATION OF LINEAR MOMENTUM Initial momentum = Final momentum

Total momentum before impact = Total momentum after impact

m1u1 + m2u2 = m1v1+m2v2

52. IMPACT An sudden short term action between two bodies

53. ELASTIC BODY If body retains its original shape and size when the

external forces are removed, then the body is said to be perfectly elastic body.

54. PERIOD OF DEFORMATION

The time elapsed from the instant of initial contact to the

maximum deformation is period of deformation.

55. PERIOD OF RESTITUTION The time elapsed from the instant of maximum deformation

to the instant of just separation is called period of restitution.

period of impact = period of deformation + period of restitution

56. NEWTON’S LAW OF COLLISTION (CO-EFFICIENT OF

RESTITUTION) Coefficient of restitution (e) = Impulse during restitution

Impulse during deformation

= V2-V1

u1-u2

57. IMPACT OF ELASTIC BODY WITH A FIXED PLANE Coefficient of restitution = Velocity of restitution

Velocity of deformation

e = - (2gh) ½

(2gH) ½

h = e²H

58. LINE OF IMPACT It is an imaginary line passing through the point of

contact and normal to the plane of contact

59. DIRECT IMPACT The velocities of two colliding bodies before impact

and after impact are collinear with the line of impact

60. OBLIQUE IMPACT The velocities of two collding bodies after collision are not

collinear with the line of impact

Vertical component before impact = vertical component after impact

Horizontal component before impact = horizandal component after impact

MODULE V

FRICTION AND ELEMENTS OF RIGID BODY DYNAMICS

1) Friction

The tangential force developed in opposite direction of

the movement of object is called force of friction or frictional

force or simply friction.

Types of friction

1. Dry friction ����absence of lubricating oil.

i. Static friction ����friction of body under rest.

ii. Dynamic (Kinetic) friction ����friction of body

under motion.

2. Fluid friction ����presence of thin film of fluid.

2) Limiting friction

The maximum resistance offered by a body against the

external force which tends to move the body is called limiting

force of friction.

3) Coefficient of friction (µµµµ)

µ=Fm

NR

Fm � Force of friction.

NR � Normal reaction.

4) For static friction

(Fm) s=µs NR > (Fm) k

For dynamic friction

(Fm) k=µk NR

5) Reaction of friction

R=√(NR2+Fm

2)

6) Angle of friction (φφφφ)

�angle between normal reaction and reaction is called

angle of friction.

tan φ=Fm =µ

NR

7) If a body is under equilibrium under the action of W &

NR

Case (i) F=0 , W=NR

Case (ii) F≤Fm , W=NR , F=Fm

Case (iii) F=Fm, W=NR , Fm =µNR

Case (iv) F>Fm (body gets movement) ,W=NR, F=Fm,

Fm=µNR

8) Impending motion

The state of motion of a body which is just about to move

or slide is called impending motion.

9) Coulomp’s law of static friction

� Fm opposite to direction of P (applied force)

� Fm independent of shape of body and area of contact

� Fm depends on degree of roughness of surface

� Fm=P when body is at rest

� (Fm)s=µs NR

10) Coulomp’s law of dynamic or kinetic friction

� Fm opposite to direction of P

� (Fm)k<(Fm)s

� µk<µs

� (Fm)k=µk NR

11) Angle of repose (ααααm)

Angle of inclined plane at which the body tends to slide

down is known as angle of repose.

µ=tan αm=tan φ

αm=φ at the time of repose.

12) Simple contact friction

The type of friction between the surface of block and plane

is called simple contact friction.

13) Applications of simple contact friction

� Ladder friction

� Screw friction

� Belt friction

14) Screw jack

Screw jack is a device used to lift or lower loads

gradually.

15) Screw friction

Friction in screw jack.

16) Force applied at lever

� While lifting P1=Wr tan(φ+α)

L

� While lowering P1=Wr tan(φ-α)

L

17) Pitch

Pitch (p) is the distance moved by the screw in one

cycle rotation.

18) Statically determinate structure

If a body or structure can be analysed using

equilibrium equations ΣH=0, ΣV=0 & ΣM=0 then it is said to

be statically determinate structure.

19) Force transmitted on screw

• To lift

P=W tan (φ+α)

• To lower

P=W tan (φ-α) α→ helical angle

φ→ angle of friction

20) Screw friction

Helical angle α=tan-1

p

21)

πd

Angle of friction φ ==== tan-1

Fm

NR

p→pitch

d→diameter Fm→Friction force

NR→Normal reaction

21) Belt friction

T2

= eµθ

T1

T1→Tension in slack side

T2→Tension in tight side

µ→Coefficient of friction between belt and wheel

θ→Angle of contact

22) Power transmitted

P=(T2-T1)×V Nm/s (watt) V→velocity of belt

23) Torque on driver=(T2-T1)×r1 N-m

Torque on follower=(T2-T1)×r2 N-m

24) Rolling resistance

When a body is made to roll over another body the

resistance developed in the opposite direction of motion is

called the rolling resistance. Rcosα is called rolling resistance.

25) Coefficient of resistance

Horizontal distance b is called as coefficient of

resistance.

b=Pr

W

P→applied force

r→radius of the body

W→weight of the body

26) Types of motion of rigid bodies

� Translation

a) Rectilinear→ straight-line path motion

b) Curvilinear→ curved path motion

� Rotation (with respect to a fixed point)

� Rotation & Translation (General plane method)

27) Angular displacement (θθθθ)))) →Total angle which a body has rotated in t

seconds.

θ=f(t) rad

28) Angular velocity (ωωωω)))) → Rate of change of angular displacement.

ω=dθ rad/s

dt

29) Angular acceleration (αααα)))) →→→→Rate of change of angular velocity.

α=dω =d2θ rad/s

2

dt dt2

30) Equations of Angular motion

� θ=ωοt+1αt2

2

� ω=ωο+αt

� ω2=ωο2+2αθ

31) Relation between linear and angular motion

♠ s=rθ

♠ v=rω

♠ a=rα

32) If angular velocity (ωωωω) is uniform then angular

acceleration (αααα)=0

Hence θ=ωt

33) Angular velocity (ωωωω) is represented as rad/s or rev/min

(rpm)

ω=2πΝ

60

Ν→ rpm of rotating body.